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second word

n. bits. first word. second word. • • •. i. th word. • • •. last word. Figure 2.5. Memory words. 32 bits. • • •. b. b. b. b. 31. 30. 1. 0. for positive numbers. Sign bit:. b. =. 0. 31. for negative numbers. b. =. 1. 31. (a) A signed integer. 8 bits.

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second word

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  1. n bits first word second word • • • i th word • • • last word Figure 2.5. Memory words.

  2. 32 bits • • • b b b b 31 30 1 0 for positive numbers Sign bit: b = 0 31 for negative numbers b = 1 31 (a) A signed integer 8 bits 8 bits 8 bits 8 bits ASCII ASCII ASCII ASCII character character character character (b) Four characters Figure 2.6.Examples of encoded information in a 32-bit word.

  3. Address Contents i Begin execution here Move A,R0 3-instruction i + 4 program Add B,R0 segment i + 8 Move R0,C A Data for B the program C Figure 2.8. A program for C ¬ [A] + [B].

  4. Move NUM1,R0 i i + 4 Add NUM2,R0 i + 8 Add NUM3,R0 • • • i Add NUM n ,R0 + 4 n - 4 i 4 n + Move R0,SUM • • • SUM NUM1 NUM2 • • • NUM n Figure 2.9. A straight-line program for addingn numbers.

  5. Move N,R1 Clear R0 LOOP Determine address of "Next" number and add "Next" number to R0 Program loop Decrement R1 Branch>0 LOOP Move R0,SUM • • • SUM N n NUM1 NUM2 • • • NUM n Figure 2.10. Using a loop to addn numbers.

  6. Add (R1),R0 Add (A),R0 Main memory B Operand A B Register Operand R1 B B (a) Through a general-purpose register (b) Through a memory location Figure 2.11. Indirect addressing.

  7. Address Contents Move N,R1 Initialization Move #NUM1,R2 Clear R0 LOOP Add (R2),R0 Add #4,R2 Decrement R1 LOOP Branch>0 Move R0,SUM Figure 2.12. Use of indirect addressing in the program of Figure 2.10.

  8. Add 20(R1),R2 R1 1000 1000 20 = offset 1020 Operand (a) Offset is given as a constant Add 1000(R1),R2 1000 20 R1 20 = offset 1020 Operand (b) Offset is in the index register Figure 2.13. Indexed addressing.

  9. n N Student ID LIST LIST + 4 Test 1 Student 1 LIST + 8 Test 2 LIST + 12 Test 3 LIST + 16 Student ID Test 1 Student 2 Test 2 Test 3 • • • Figure 2.14. A list of students' marks.

  10. Move #LIST,R0 Clear R1 Clear R2 Clear R3 Move N,R4 Add 4(R0),R1 LOOP Add 8(R0),R2 A d d 12(R0),R3 Add #16,R0 Decrement R4 Branch>0 LOOP Move R1,SUM1 R2,SUM2 Move Move R3,SUM3 Figure 2.15. Indexed addressing used in accessing test scores in the list in Figure 2.14.

  11. Move N,R1 Initialization Move #NUM1,R2 Clear R0 LOOP Add (R2)+,R0 Decrement R1 Branch>0 LOOP Move R0,SUM Figure 2.16. The Autoincrement addressing mode used in the program of Figure 2.12.

  12. 100 Move N,R1 104 Move #NUM1,R2 108 Clear R0 LOOP 112 Add (R2),R0 116 Add #4,R2 120 Decrement R1 124 Branch>0 LOOP 128 Move R0,SUM 132 SUM 200 N 204 100 NUM1 208 NUM2 212 NUM n 604 Figure 2.17. Memory arrangement for the program in Figure 2.12.

  13. Memory Addressing address or data lab el Op eration information Assem bler directiv es SUM EQU 200 ORIGIN 204 N D A T A W ORD 100 NUM1 RESER VE 400 ORIGIN 100 Statemen ts that ST AR T MO VE N,R1 generate MO VE #NUM1,R2 mac hine CLR R0 instructions LOOP ADD (R2),R0 ADD #4,R2 DEC R1 BGTZ LOOP MO VE R0,SUM Assem bler directiv es RETURN END ST AR T Figure 2.18. Assembly language representation for the program in Figure 2.17.

  14. Mo v e #LOC,R0 Initialize p oin ter register R0 to p oin t to the address of the first lo cation in memory where the c haracters are to b e stored. READ T estBit #3,INST A TUS W ait for a c haracter to b e en tered Branc h=0 READ in the k eyb oard buffer D A T AIN. Mo v eByte D A T AIN,(R0) T ransfer the c haracter from D A T AIN in to the memory (this clears SIN to 0). ECHO T estBit #3,OUTST A TUS W ait for the displa y to b ecome ready . Branc h=0 ECHO Mo v eByte (R0),D A T A OUT Mo v e the c haracter just read to the displa y buffer register (this clears SOUT to 0). Compare #CR,(R0)+ Chec k if the c haracter just read is CR (carriage return). If it is not CR, then Branc h READ branc h bac k and read another c haracter. 0 Also, incremen t the p oin ter to store the next c haracter. Figure 2.20. A program that reads a line of characters and displays it.

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